$p(t) = (t, e^t, t)$ What is the speed of $p(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(0, e^t, 0)$ (Choice B) B $\sqrt{2 + e^{t^2}}$ (Choice C) C $(1, e^t, 1)$ (Choice D) D $\sqrt{2 + e^{2t}}$
The speed of a parametric curve is the magnitude of its velocity. If $f(t) = (a(t), b(t), c(t))$, then speed is: $\| f'(t) \| = \sqrt{ a'(t)^2 + b'(t)^2 + c'(t)^2 }$ Our position function here is $p(t)$. $\begin{aligned} p'(t) &= (1, e^t, 1) \\ \\ \text{speed} &= ||p'(t)|| \\ \\ &= \sqrt{1 + e^{2t} + 1} \\ \\ &= \sqrt{2 + e^{2t}} \end{aligned}$ Therefore, the speed of $p(t)$ is $\sqrt{2 + e^{2t}}$.